Question: You have found the following ages (in years) of 4 bears. Those bears were randomly selected from the 22 bears at your local zoo: $ 5,\enspace 4,\enspace 6,\enspace 39$ Based on your sample, what is the average age of the bears? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 22 bears, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{5 + 4 + 6 + 39}{{4}} = {13.5\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {72.25} + {90.25} + {56.25} + {650.25}} {{4 - 1}} $ {s^2} = \dfrac{{869}}{{3}} = {289.67\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{289.67\text{ years}^2}} = {17\text{ years}} $ We can estimate that the average bear at the zoo is 13.5 years old. There is also a standard deviation of 17 years.